Question: The Fibonacci sequence is defined $F_1 = F_2 = 1$ and $F_n = F_{n - 1} + F_{n - 2}$ for all $n \ge 3.$

The Fibonacci numbers $F_a,$ $F_b,$ $F_c$ form an increasing arithmetic sequence.  If $a + b + c = 2000,$ compute $a.$
Solution: We claim that if $F_a,$ $F_b,$ $F_c$ form an increasing arithmetic sequence, then $(a,b,c)$ must be of the form $(n,n + 2,n + 3)$ for some positive integer $n.$  (The only exception is $(2,3,4).$)

From $F_c - F_b = F_b - F_a,$ we get
\[F_c = F_b + (F_b - F_a) < F_b + F_{b + 1} = F_{b + 2}.\]Also, $F_c > F_b.$  Therefore, $F_c = F_{b + 1}.$

Then
\begin{align*}
F_a &= 2F_b - F_c \\
&= 2F_b - F_{b + 1} \\
&= F_b - (F_{b + 1} - F_b) \\
&= F_b - F_{b - 1} \\
&= F_{b - 2}.
\end{align*}Then $a$ must be equal to $b - 2$ (unless $b = 3,$ which leads to the exceptional case of $(2,3,4)$).  Taking $n = b - 2,$ we get $(a,b,c) = (n,n + 2,n + 3).$

Then $a + (a + 2) + (a + 3) = 2000,$ so $a = \boxed{665}.$